//给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。 
//
// 叶子节点 是指没有子节点的节点。 
//
// 
//
// 示例 1： 
//
// 
//
// 
//输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
//输出：[[5,4,11,2],[5,8,4,5]]
// 
//
// 示例 2： 
//
// 
//
// 
//输入：root = [1,2,3], targetSum = 5
//输出：[]
// 
//
// 示例 3： 
//
// 
//输入：root = [1,2], targetSum = 0
//输出：[]
// 
//
// 
//
// 提示： 
//
// 
// 树中节点总数在范围 [0, 5000] 内 
// -1000 <= Node.val <= 1000 
// -1000 <= targetSum <= 1000 
// 
//
// 注意：本题与主站 113 题相同：https://leetcode-cn.com/problems/path-sum-ii/ 
// Related Topics 树 深度优先搜索 回溯 二叉树 👍 337 👎 0

package leetcode.editor.offer;

import java.util.LinkedList;
import java.util.List;

// 34. 二叉树中和为某一值的路径
// https://leetcode.cn/problems/er-cha-shu-zhong-he-wei-mou-yi-zhi-de-lu-jing-lcof/submissions/
class ErChaShuZhongHeWeiMouYiZhiDeLuJingLcof {
    public static void main(String[] args) {
        Solution solution = new ErChaShuZhongHeWeiMouYiZhiDeLuJingLcof().new Solution();
    }
    //leetcode submit region begin(Prohibit modification and deletion)

    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode() {
        }

        TreeNode(int val) {
            this.val = val;
        }

        TreeNode(int val, TreeNode left, TreeNode right) {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }

    class Solution {
        // 方法一
        /*public List<List<Integer>> pathSum(TreeNode root, int target) {
            if (root == null) {
                return res;
            }
            int count = target - root.val;
            dfs(root, path, count);
            return res;
        }

        List<List<Integer>> res = new ArrayList<>();
        LinkedList<Integer> path = new LinkedList<>();

        public void dfs(TreeNode root, LinkedList<Integer> path, int count) {
            path.add(root.val);
            if (root.left == null && root.right == null) {
                if (count == 0) {
                    res.add(new LinkedList<>(path));
                }
                return;
            }

            if (root.left != null) {
                count -= root.left.val;
                dfs(root.left, path, count);
                count += root.left.val;    // 回溯
                path.removeLast();
            }

            if (root.right != null) {
                count -= root.right.val;
                dfs(root.right, path, count);
                count += root.right.val;    // 回溯
                path.removeLast();
            }

            return;
        }*/

        // 方法二

        /*LinkedList<List<Integer>> res = new LinkedList<>();
        LinkedList<Integer> path = new LinkedList<>();

        public List<List<Integer>> pathSum(TreeNode root, int target) {
            recur(root, target);
            return res;
        }

        public void recur(TreeNode root, int tar) {
            if (root == null) {
                return;
            }
            path.add(root.val);
            tar -= root.val;

            // 注意判断左右孩子
            if (tar == 0 && root.left == null && root.right == null) {
                res.add(new LinkedList(path));
            }

            recur(root.left, tar);
            recur(root.right, tar);
            path.removeLast();
        }*/

        LinkedList<List<Integer>> res = new LinkedList<>();
        LinkedList<Integer> path = new LinkedList<>();

        public List<List<Integer>> pathSum(TreeNode root, int target) {
            if (root == null) return res;
            path.add(root.val);
            dfs(root, target - root.val);
            return res;
        }

        public void dfs(TreeNode root, int target) {
            if (root.left == null && root.right == null) {
                if (target == 0) {
                    res.add(new LinkedList<>(path));
                }
                return;
            }

            if (root.left != null) {
                path.add(root.left.val);
                dfs(root.left, target - root.left.val);
                path.removeLast();
            }


            if (root.right != null) {
                path.add(root.right.val);
                dfs(root.right, target - root.right.val);
                path.removeLast();
            }

        }
    }
//leetcode submit region end(Prohibit modification and deletion)

}
